# ap calculus

posted by on .

The height s of a ball, in feet, thrown with an initial velocity of 112 feet per second from an initial height of 6 feet is given as a function of time t (in seconds) by
s(t)=-16t^2+112t+20.

1. What is the maximum height of the ball?

2. At what time does the maximum height occur?

I know how to find the maximum height using the formula -b/2a and I got 3.5. Usually when I find the maximum height I would plug 3.5 into the original equation. The problem says initial height of 6 feet which is not in the equation presented to me so I'm confused about how to solve for the time if the 6 feet is supposed to be somewhere.

• pre-calc - ,

oops! don't know why I put AP Calc, it's actually pre-calc....

• ap calculus - ,

I am baffled by the same question. You may want to double check the numbers, or your instructor could help you.

As you suggested, if there is no typo, then either the reference point for the height is 14 feet below the mentioned datum, or the time reference is shifted.

• ap calculus - ,

Thanks for the info. I will ask my teacher on Monday.

• ap calculus :) - ,

You're welcome.
I hope your teacher will clear this up on Monday.